Let $\mu$ denote the Bernoulli measure on $\Sigma_k^+$ determined by $(\frac{1}k, \ldots , \frac{1}{k})$. Show that, for each $k\ge 2$, $\sigma_k$ (with respect to $\mu$) and $T_k$ (with respect to Lebesgue measure) are measure-theoretically isomorphic.
If we take the symbol set for $\Sigma_k^+$ to be $\{0, 1, . . . , k − 1\}$, and let $M_k$ be the set of sequences which do not end in an infinite string of zeros. Then $\mu(M_k) = 1$ and $M_k$ is preserved by the shift map.
Now, the isomorphism need only be defined a.e., so define a map $\phi : M_k → \mathbb{R}/\mathbb{Z}$ by
$$ \phi(x_0,x_1,\ldots)= \sum_{j=0}^\infty \frac{x_j}{k^{j+1}} \quad \text{mod 1}$$
I am not sure, but would the above method create a well-defined bijection?
If so then $\phi^{-1}\left(\frac{i}{k^j},\frac{i + 1}{k^j}\right)$ is a length $k$ cylinder and if $C \subset \Sigma_k^+$ is a length n cylinder then $\phi(C)$ is a k-adic interval of length $k^{-n}$. This shows that $\phi$ and $\phi^{-1}$ are measurable and measure-preserving with respect to $\mu$ and $\lambda$.
I then get the isomporphism:
\begin{align} \phi \circ \sigma_k(x_0, x_1, \ldots) &= \phi(x_1, x_2, \ldots) \\ &=\sum_{j=0}^\infty \frac{x_{j+1}}{k^{j+1}}\\ &=\sum_{j=1}^\infty \frac{x_j}{k^{j}}\\ &=\sum_{j=0}^\infty \frac{kx_j}{k^{j+1}}\\ &= k\phi(x_0, x_1, \ldots) \quad \text{mod 1} \end{align}
Can I check this is correct?